Phy 231
Your 'torques' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your optional message or comment: **
** Positions of the three points of application, lengths of systems B, A and C (left to right), the forces in Newtons exerted by those systems, description of the reference point: **
.75, 7.6, 14.65
8.1, 8.05, 8.6
1.045, 2.9133, 1.90
I used the point that was the left end of the rod as point 0.
I used proportions from the known force and distance values.
In the first line I gave in the order of B, A, C the distances in centimeters from the left of the rod the points where the rubber band systems exerted force.
** Net force and net force as a percent of the sum of the magnitudes of all forces: **
-0.0317
0.617
What my results mean is that I set the rubber band systems B, A, and C to have a net force of 0.0317 N in the downwards direction, which is only .617% of the total force being applied to the rod by the rubber band systems.
** Moment arms for rubber band systems B and C **
-6.85, 7.05
This indicates the distance in centimeters that the forces of rubber band system B and then rubber band system C are acting from the fulcrum point located at the point where rubber band system A exerted force. The negative sign for the moment-arm of system B indicates the force is exerted in the counter-clockwise direction, and the positive sign on system A indicates the force is exerted in the clockwise direction.
** Lengths in cm of force vectors in 4 cm to 1 N scale drawing, distances from the fulcrum to points B and C. **
4.18, 11.6532, 7.6
6.85, 7.05
The first line shows the force with a 4 cm/N ratio of the three rubber band systems forces as given in centimeters. The second line first indicates how far to the left of the fulcrum point B is and the second value of this line indicates how far to the right point C is from the fulcrum.
** Torque produced by B, torque produced by C: **
7.16, -13.4
I determined these values by multiplying the distance in centimeters from the fulcrum by the perpendicular force being applied by the corresponding rubber band system; therefore, the values above are of N*cm, and not the standard torque unit of N*m.
** Net torque, net torque as percent of the sum of the magnitudes of the torques: **
-6.24
30.3
I obtained this percentage by taking the magnitude of net torque, 6.24 N*cm, and dividing this by the magnitude of the sum of the two systems of torque, which is 20.56, and then multiplied the quotient by 100.
This means that according to these values the rod should be rotating in a clockwise direction with a force roughly 1/3 of the total torque being applied by the two rubber band systems.
** Forces, distances from equilibrium and torques exerted by A, B, C, D: **
2.28, 0, 0
1.292, .9, -1.1628
.76, 12.30, -9.348
.76, 13.55, 10.30
The first column indicates the force exerted perpendicular to the rod, the second column indicates the distance in centimeters from the leftmost force acting upon the rod, and the third column indicates the torque by each rubber band system as measured in N*cm.
** The sum of the vertical forces on the rod, and your discussion of the extent to which your picture fails to accurately describe the forces: **
.998
My picture includes two upward force vector at positions of 0 cm and 13.55 cm with 2.28 N and .76 N correspondingly. It also shows two downward forces at positions .9 cm and 12.3 cm with 1.292 N and .76 N correspondingly.
** Net torque for given picture; your discussion of whether this figure could be accurate for a stationary rod: **
- 0.2108
This shows that the rod should have net torque in the clockwise direction, which is the direction the rod would have turned; however, since the rod is not in motion and is stationary it is not accurate to state their is any net force or torque acting upon the rod.
** For first setup: Sum of torques for your setup; magnitude of resultant and sum of magnitudes of forces; magnitude of resultant as percent of sum of magnitudes of forces; magnitude of resultant torque, sum of magnitudes of torques, magnitude of resultant torque as percent of the sum of the magnitudes: **
-.2108
.998, 5.092
19.6
.2108, 20.81, 1.01
The first line gives the net torque and direction of this system. The second line shows the net force and direction acting upon this system, and this sum of the magnitude of all of these forces. The third line gives the percentage the net force is of the total force given previously. The fourth line gives the magnitude of net torque, the sum of all torques, and the percentage relating the net torque to the sum of torques. All force values are measured in Newtons, and all torque values are measured in N*cm.
** For second setup: Sum of torques for your setup; magnitude of resultant and sum of magnitudes of forces; magnitude of resultant as percent of sum of magnitudes of forces; magnitude of resultant torque, sum of magnitudes of torques, magnitude of resultant torque as percent of the sum of the magnitudes: **
.4997
.998, 6.232
16.01
.4997, 35.54, 1.41
The first line gives the net torque and direction of this system. The second line shows the net force and direction acting upon this system, and this sum of the magnitude of all of these forces. The third line gives the percentage the net force is of the total force given previously. The fourth line gives the magnitude of net torque, the sum of all torques, and the percentage relating the net torque to the sum of torques. All force values are measured in Newtons, and all torque values are measured in N*cm.
** In the second setup, were the forces all parallel to one another? **
No the forces were not parallel to one another. The forces of the rubber band systems C and D were acting in a direction approximately 1 degree different from the forces of the rubber band systems A and B. I made my estimate by observing that the forces of the A and B systems were parallel to the lines of the graph, but the forces of the C and D system were at an angle relative to the rod that would be given as 269 degrees which is 2 degrees different from the 270 degrees acting downwards in direction, but perpendicular to the rod that the A and B system have.
** Estimated angles of the four forces; short discussion of accuracy of estimates. **
90, 270, 269, 90
I believe my estimates to be very accurate, because the relative change from a perpendicular position was very small, which means that relative to 90 being off by a single degree would not have significant influence on any following calculations.
** x and y coordinates of both ends of each rubber band, in cm **
1.9, 2.5, 1.85, 10.65
6.7, 16.8, 5.2, 24.85
14.9, 4.8, 12.2, 12.5
In order to accurately measure the distances from of the coordinates I drew lines from the coordinate points that would intersect the x and y axis perpendicularly and allow me to measure with minimal error.
** Lengths and forces exerted systems B, A and C:. **
8.15, 1.14
8.16, 1.72
8.19, 3.38
The lengths were all found by the Pythagorean Theorem as detailed in the above instructions. The force values were found for the corresponding rubber calibration values from the previous rubber band calibration lab.
** Sines and cosines of systems B, A and C: **
-1, 0
.983, -0.183
-.944, -0.331
The values in the first column represent the sine values determined by the change in rise of the coordinates divided the distance between the two coordinates or the hypotenuse of the triangle.
** Magnitude, angle with horizontal and angle in the plane for each force: **
1.14, 270.00, 259.68
3.38, 100.56, 90.14
1.72, 287.99, 277.57
The first column is the magnitude of the force exerted by each system. The values in the second column represent the angle made by the x-axis and the hypotenuse of each triangle as found by the instructions above. The third column represents the angle found relative the plane of the rod, which is elevated 10.42 degrees above that of the x-axis of the grid.
** x and y components of sketch, x and y components of force from sketch components, x and y components from magnitude, sine and cosine (lines in order B, A, C): **
-0.7, -4.45, -.175, -1.11, -.204, -1.12
-.25, 13.55, -.0625, 3.39, -.00826, 3.38
1.2, -7.5, 0.3, -1.88, 0.236, -1.77
The values in the first, third, and fifth column show the x-components of the force as measured in centimeter by sketch, Newton centimeter by sketch, and Newton centimeter by calculation. The second, fourth, and sixth columns show the same for the y-components.
** Sum of x components, ideal sum, how close are you to the ideal; then the same for y components. **
.0625, .0237, .0388
.4, .49, .09
I obtained the number in the first columns by adding the third and fifth, respectively to find the values here. I obtained the number in the second column by adding the fourth and sixth columns above respectively to find the values here. The last column is the difference between the first and second column here.
** Distance of the point of action from that of the leftmost force, component perpendicular to the rod, and torque for each force: **
0, -1.12, 0
5.45, 3.38, 18.42
9.75, -1.77, -17.26
I obtained the values in the first column by measuring the distance from the point of force exerted by system B to the next point of force exerted in centimeters. The second column was obtained previously and repeated here. The third column is the product of the values in the corresponding first and second columns. This shows that the system have greater torque in the positive counter-clockwise direction than it did in the negative clockwise direction.
** Sum of torques, ideal sum, how close are you to the ideal. **
1.16, 0, 1.16
The first value is simply the sum of 18.42 + (-17.26) = 1.16. The second column is 0, because since the rod is not moving the ideal net torque should be 0. My value was therefore different by 1.16 - 0 = 1.16. All values here are measured in N * cm.
** How long did it take you to complete this experiment? **
3 hours and 30 minutes
** Optional additional comments and/or questions: **
Excellent work.